Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.
When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes \(\frac{1}{6}\) hours per mile, Butch takes \(\frac{1}{4}\) hours per mile, and Sundance takes \(\frac{2}{5}\) hours per mile.
Time
Distance
Speed
Answer: is 279.
AIME, 2012, Question 4
Problem Solving Strategies by Arther Engel
After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.
Then
\(\frac{a}{6} + \frac{n-a}{4}\) = \(\frac{n-a}{6} + \frac{2a}{5}\) implies that \(a = \frac{5}{19}n\)
Then integral value of n is 19 and a = 5 and \(t = \frac{13}{3}\) hours that is 260 minutes. Then \(19 + 260 = {279}\).

In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]

Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.

Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.

Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.